Question

The mean number of hours per day spent watching television, according to a national survey, is 3.5 hours, with a standard deviation of two hours. If each time was increased by one hour, what would be the new mean and standard deviation? (2 points)

Answer 1

Answer:

4.5 hour and 2 hour

Step-by-step explanation:

Given: The mean number of hours per day spent watching television, according to a national survey, is 3.5 hours, with a standard deviation of two hours.

To Find: If each time was increased by one hour, what would be the new mean and standard deviation.

Solution:

let the total numbers entries of hours in survey be   =   $\text{N}$

each entry in survey be  =  $\text{x}_{i}$

mean of survey is     =     $\mu$  =$3.5$ $\text{hours}$

standard deviation is  =   $\sigma$ = $2$ $\text{hours}$

if each entry in survey is increased by one hour then,

each new entry in survey  is =  $\text{x}_{i}+1$

the new mean is$\mu_{new}$  = $\frac{\text{sum of all hours}}{total number of entries}$

                                                $\frac{\text{x}_{1} +1+\text{x}_{2}+1 +....+\text{x}_{N}+1 }{N}$

                                                $\frac{(\text{x}_{1} +\text{x}_{2} +....+\text{x}_{N})+\text{N}\times1 }{N}$

                                                $\frac{(\text{x}_{1} +\text{x}_{2} +....+\text{x}_{N})}{N}$+$\frac{\text{N}\times1 }{N}$

                                             $\mu_{new}$ = $\mu$ + 1=$4.5$ $\text{hours}$

now,

standard deviation is      $\sigma_{new}=$ $\sqrt{\sum_{1}^{N}\frac{(\text{x}_{inew}-\mu_{new})^{2}}{N}}$

                                   $\text{x}_{inew}= \text{x}_{i}+1$

                                   $\mu_{new}=\mu+1$        

putting values,

                                                $\sqrt{\sum_{1}^{N}\frac{(\text{x}_{i}+1-\mu-1)^{2}}{N}}$

                                                $\sqrt{\sum_{1}^{N}\frac{(\text{x}_{i}-\mu)^{2}}{N}}$= $\sigma$

                                 $\sigma_{new}$        =2

new mean and standard deviation are $4.5$ and $2$ $\text{hours}$